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Bruno Stonek
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Why isn't the perfect closure separable?

This question has escalated from math.stackexchange. I'm doing this because it has been a while since the question has been open, receiving no satisfactory answers, even when subject to a bounty. I hope it is adequate for MO, I apologize if it is not.

Let $F\subset K$ be an algebraic extension of fields. By taking the separable closure $K_s$, we obtain a tower $F\subset K_s \subset K$ such that $F\subset K_s$ is separable and $K_s\subset K$ is purely inseparable.

Wikipedia, following Isaacs, Algebra, a graduate course p.301, says:

On the other hand, an arbitrary algebraic extension $F\subset K$ may not possess an intermediate extension $E$ that is purely inseparable over $F$ and over which $K$ is separable.

The question is: why? And more explicitly, had I not seen this this soon, I would surely have conjectured that the perfect closure, $K_p$, which satisfies $F\subset K_p$ purely inseparable, also satisfied $K_p\subset K$ separable... But why doesn't it?

To clarify, I'm looking both for a counterexample and for intuition regarding the impediment of the situation to be symmetrical.

Bruno Stonek
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